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t, Chi-Square and F Distributions · Tutorial 493 of 996

How Degrees of Freedom Shape F Curves

Learn how changing either degree of freedom reshapes an F curve and why the two degrees of freedom have distinct effects.

Intermediate 10 min read

What You'll Learn

  • Describe the support and right-skewed shape of an F distribution.
  • Compare how numerator degrees of freedom affect the mode and variability.
  • Explain how denominator degrees of freedom affect the mean and right tail.
  • Use the F density to locate its mode and justify how it changes.
  • Recognize when the mean, variance, and skewness are defined.
  • Avoid interpreting a changing peak as a uniform shift of the whole curve.

Why the Two Degrees of Freedom Matter

An F curve is determined by two degrees of freedom, but they do not play interchangeable roles. The numerator degrees of freedom affect the variation in the numerator of the ratio; the denominator degrees of freedom affect the variation in its denominator. As a result, changing one while holding the other fixed can alter the peak, spread, and right tail in different ways.

A useful way to compare F curves is to distinguish their mode (the location of the peak), their mean (when it exists), and their tail behavior. These summaries do not always move together. In particular, the mode can move right while the mean moves left, so it is misleading to describe a whole curve as simply shifting in one direction.

Definition: If \(U\sim\chi^2_{d_1}\) and \(V\sim\chi^2_{d_2}\) are independent, then \(F=(U/d_1)/(V/d_2)\) has an F distribution with numerator degrees of freedom \(d_1\) and denominator degrees of freedom \(d_2\), written \(F\sim F(d_1,d_2)\). Its values are positive. The distribution is right-skewed, with a long tail extending toward large positive values.

The model requires both scaled chi-square variables to be formed with their stated degrees of freedom and to be independent. In this tutorial, comparisons change one degree of freedom at a time so that the effect of that change can be separated from the other.

The Density and Its Main Summaries

For \(x>0\), the density of \(F(d_1,d_2)\) is

$$ f(x)= \frac{(d_1/d_2)^{d_1/2}}{B(d_1/2,d_2/2)} x^{d_1/2-1} \left(1+\frac{d_1x}{d_2}\right)^{-(d_1+d_2)/2}, $$

where \(B(a,b)\) is the beta function, a positive constant for positive \(a\) and \(b\). The density is zero for \(x\leq 0\). The density formula allows us to locate the peak, while moment formulas summarize the distribution’s center and spread when those moments exist.

Key results: For \(F\sim F(d_1,d_2)\), the mode is the boundary value \(0\) (a limiting peak) if \(d_1\leq 2\), and otherwise $$ \operatorname{mode}(F)=\frac{d_1-2}{d_1}\frac{d_2}{d_2+2}. $$ The mean exists when \(d_2>2\), and then \(E(F)=d_2/(d_2-2)\). The variance exists when \(d_2>4\), and then $$ \operatorname{Var}(F)= \frac{2d_2^2(d_1+d_2-2)} {d_1(d_2-2)^2(d_2-4)}. $$ The skewness exists when \(d_2>6\), and then $$ \operatorname{Skewness}(F)= \frac{(2d_1+d_2-2)\sqrt{8(d_2-4)}} {(d_2-6)\sqrt{d_1(d_1+d_2-2)}}. $$

The mode is a peak location, not an average. If \(d_1\leq2\), the density is largest in the limit as \(x\to0^+\), so \(0\) is the boundary (limiting) mode rather than an attained maximum: for \(d_1<2\), the density grows without bound as \(x\) approaches zero from above, and for \(d_1=2\) it approaches its largest value there. The mean and variance have stricter existence conditions because the right tail can be sufficiently heavy that the corresponding integrals do not converge.

Why the Mode Has This Formula

For \(x>0\), the factors in the density that do not depend on \(x\) cannot affect where its maximum occurs. Take the logarithm of the \(x\)-dependent part, differentiate, and set the derivative equal to zero:

$$ \begin{aligned} \log f(x)&=\text{constant}+ \left(\frac{d_1}{2}-1\right)\log x -\frac{d_1+d_2}{2}\log\left(1+\frac{d_1x}{d_2}\right),\\ \frac{d}{dx}\log f(x) &=\frac{d_1/2-1}{x} -\frac{d_1+d_2}{2}\frac{d_1/d_2}{1+d_1x/d_2}. \end{aligned} $$

When \(d_1>2\), solving this derivative equation gives

$$ \frac{d_1-2}{2x} = \frac{d_1(d_1+d_2)}{2(d_2+d_1x)} \quad\Longrightarrow\quad x=\frac{d_2(d_1-2)}{d_1(d_2+2)}. $$

The density increases before this positive critical point and decreases after it, so this point is the mode. When \(d_1\leq2\), the density is decreasing from the boundary (or has its boundary maximum), giving mode \(0\). This derivation also shows why both degrees of freedom appear in the location of the peak.

Worked Example: Finding the Mode for \(F(5,8)\)

Consider an F curve with numerator degrees of freedom \(d_1=5\) and denominator degrees of freedom \(d_2=8\). Since \(d_1>2\), its mode is positive and the mode formula applies.

Substitute and calculate:

$$ \operatorname{mode}(F) =\frac{5-2}{5}\frac{8}{8+2} =\frac{3}{5}\cdot\frac{8}{10} =0.6\cdot0.8 =0.4800. $$

The density reaches its peak at \(F=0.4800\). The mode is not the mean: because \(d_2=8>2\), the mean is \(8/(8-2)=1.333\). A peak below the mean is consistent with a long right tail pulling the mean upward.

Changing the Numerator Degrees of Freedom

Hold \(d_2\) fixed and increase \(d_1\). The numerator \(U/d_1\) becomes less variable as its degrees of freedom increase: its mean remains \(1\), while its variance is \(2/d_1\). Correspondingly, for fixed \(d_2>4\), the variance formula for the F variable decreases as \(d_1\) increases. When the mode formula applies, the factor \((d_1-2)/d_1=1-2/d_1\) increases, so the peak moves right. If \(d_2>2\), the mean \(d_2/(d_2-2)\) does not change at all with \(d_1\).

These statements describe different features, not a rigid movement of every part of the curve. For \(d_2>6\), the skewness formula also shows how skewness changes with \(d_1\); it decreases toward a positive limiting value as \(d_1\) grows. Increasing numerator degrees of freedom therefore does not make the right tail disappear.

Worked Example: Comparing Numerator Degrees of Freedom

Compare \(F(5,10)\) and \(F(20,10)\). The denominator degrees of freedom are held at \(10\), so both means exist and are the same. Both variances also exist because \(10>4\).

Means: For either curve,

$$ E(F)=\frac{10}{10-2}=\frac{10}{8}=1.250. $$

Modes: The formulas give

$$ \operatorname{mode}(F(5,10)) =\frac{5-2}{5}\frac{10}{12} =0.6\cdot0.8333 =0.5000, $$ $$ \operatorname{mode}(F(20,10)) =\frac{20-2}{20}\frac{10}{12} =0.9\cdot0.8333 =0.7500. $$

Variances: Substituting \(d_1=5\) and \(d_1=20\) separately gives

$$ \operatorname{Var}(F(5,10)) =\frac{2(10)^2(5+10-2)} {5(10-2)^2(10-4)} =\frac{2600}{1920} \approx1.354, $$ $$ \operatorname{Var}(F(20,10)) =\frac{2(10)^2(20+10-2)} {20(10-2)^2(10-4)} =\frac{5600}{7680} \approx0.7292. $$

The larger numerator degrees of freedom produce a higher mode and smaller variance, while the mean remains \(1.250\). Thus it is not correct to say that the whole curve simply shifts right: the peak moves right, but the mean stays fixed and the distribution becomes less variable.

Changing the Denominator Degrees of Freedom

Now hold \(d_1\) fixed and increase \(d_2\). The denominator \(V/d_2\) has mean \(1\) and variance \(2/d_2\), so it becomes less variable. The F mean, when it exists, is \(d_2/(d_2-2)\), which decreases toward \(1\) as \(d_2\) grows. For \(d_1>2\), the mode also increases toward \((d_1-2)/d_1\), which is below \(1\). Again, these two summaries can move in opposite directions.

The right tail is especially sensitive to denominator degrees of freedom. From the density formula, for large \(x\),

$$ f(x)\ \text{is proportional to}\ x^{-(d_2/2+1)}. $$

The exponent depends on \(d_2\), not \(d_1\). Increasing \(d_2\) makes the asymptotic density tail decay more rapidly. It does not mean the density is lower at every value of \(x\): two curves may cross. It means that sufficiently far into the right tail, the curve with larger denominator degrees of freedom has the steeper power-law decay.

Worked Example: Comparing Denominator Degrees of Freedom

Compare \(F(5,8)\) with \(F(5,20)\). The numerator degrees of freedom remain fixed at \(5\); both curves have a positive mode. Since both denominator degrees of freedom exceed \(4\), the means and variances exist.

Modes:

$$ \operatorname{mode}(F(5,8)) =\frac{3}{5}\frac{8}{10} =0.4800, \qquad \operatorname{mode}(F(5,20)) =\frac{3}{5}\frac{20}{22} \approx0.5455. $$

Means:

$$ E(F(5,8))=\frac{8}{6}\approx1.333, \qquad E(F(5,20))=\frac{20}{18}\approx1.111. $$

Variances:

$$ \operatorname{Var}(F(5,8)) =\frac{2(8)^2(5+8-2)} {5(8-2)^2(8-4)} =\frac{1408}{720} \approx1.956, $$ $$ \operatorname{Var}(F(5,20)) =\frac{2(20)^2(5+20-2)} {5(20-2)^2(20-4)} =\frac{18400}{25920} \approx0.7099. $$

The mode rises from \(0.4800\) to about \(0.5455\), but the mean falls from \(1.333\) to \(1.111\), and the variance falls substantially. The asymptotic density powers are \(x^{-5}\) for \(F(5,8)\) and \(x^{-11}\) for \(F(5,20)\), since \(8/2+1=5\) and \(20/2+1=11\). The latter therefore has the faster-decaying far-right tail.

Reading the Shape Without Overclaiming

The comparisons can be summarized, with the relevant conditions kept in view:

Change while holding the other degree fixedEffect on the modeEffect on the mean and tail
Increase \(d_1\), with \(d_1>2\)Moves the mode right toward \(d_2/(d_2+2)\)If \(d_2>2\), the mean is unchanged; if \(d_2>4\), variance decreases. The asymptotic tail power is unchanged.
Increase \(d_2\), with \(d_1>2\)Moves the mode right toward \((d_1-2)/d_1\)The mean decreases toward \(1\) when defined; the far-right density decays with a larger power.

Both degrees of freedom influence the overall density, and neither the mode nor a single moment fully describes the curve. In particular, when a mean or variance condition fails, the corresponding summary is not a finite number; do not substitute into its formula outside its stated range.

Common Mistake: Treating the Degrees of Freedom as Interchangeable

A common mistake is to assume that increasing either degree of freedom has the same effect, or to infer that a curve whose mode moves right has also moved right everywhere. The degrees of freedom occupy different positions in the ratio: \(d_1\) controls the scaled numerator, while \(d_2\) controls the scaled denominator. For example, increasing \(d_1\) with \(d_2\) fixed leaves the mean unchanged when it exists, whereas increasing \(d_2\) changes that mean. Also, a rising mode does not prevent the mean from falling, as the \(F(5,8)\) and \(F(5,20)\) comparison showed.

Key takeaway: To compare F curves, state which degree of freedom changes and which remains fixed. Then distinguish peak location, mean, variability, and far-right tail; they are separate features and need not move in the same direction.

Check Your Understanding

Use the definitions and conditions in this tutorial to answer each question.

  1. For \(F(4,12)\), calculate the mode and mean. State which formula conditions are satisfied.
  2. With \(d_2=10\) fixed, what happens to the mean and mode when \(d_1\) increases from \(5\) to \(15\)?
  3. For fixed \(d_1=6\), compare the asymptotic density-tail powers for \(F(6,8)\) and \(F(6,16)\).
  4. Why is it incorrect to conclude that every part of an F curve moves right when its mode increases?
  5. What is the condition on \(d_2\) for the variance of \(F(d_1,d_2)\) to exist?

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